10th Maths 13.2

Chapter 13 Surface Areas And Volume

Exercise 13.2

Unless stated otherwise, take

$pi = 22/7$ .

Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to

$text{1 cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of

$pi$ .

$text{Sol. :}$

$text{Given :}$

$text{Radius of Cone & hemisphere } r = 1text{ cm}$

$text{Height of Cone & hemisphere } h = r =1text{ cm}$

$text{To find :}$

$text{Volume of the solid}$

$text{Solve :}$

$text{Volume of Solid = Volume of hemisphere}$

$text{ + Volume of Cone}$

$text{V} = 2/3pir^3 + 1/3pir^2h$

$text{ }= pir^2(2/3r + 1/3h)$

$text{ }= pi(1)^2(2/3times1 + 1/3 times 1)$

$text{ }= pi(2/3 + 1/3)$

$text{ }= pi((2+1)/3)$

$text{ }= pi(3/3)$

$text{ }= pi text{ cm}^3$

$text{Answer :}$

$text{Volume of the solid is } pi text{ cm}^3.$

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is

$text{3 cm}$ and its length is

$text{12 cm}$ . If each cone has a height of

$text{2 cm}$ , find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

$text{Sol. :}$

$text{Given :}$

$text{Height of each Cone }h = 2 text{ cm}$

$text{Height of Cylinder } H = 12-(2+2)=8text{ cm}$

$text{Radius of Cone & Cylinder }r = 3/2text{ cm}$

$text{To find :}$

$text{Volume of the model}$

$text{Solve :}$

$text{Volume of model = Volume of Cylinder}$

$text{ + }2times text {Volume of Cone}$

$text{V }= pir^2H + 2times1/3pir^2h$

$text{ }= pir^2(H + 2times1/3rh)$

$text{ }= 22/7times(3/2)^2(8 + 2/3times2)$

$text{ }= 22/7times3/2times3/2(8+4/3)$

$text{ }= 11/7times3times3/2((24+4)/3)$

$text{ }= 11/7times3/2times28$

$text{ }= 11times3times2$

$text{ }= 66text{ cm}^3$

$text{Answer :}$

$text{The Volume of the air 66 cm}^3.$

Q3. A gulab jamun contains sugar syrup up to about

$30%$ of its volume. Find approximately how much syrup would be found in

$45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends with length

$text{5 cm}$ and diameter

$text{2.8 cm}$ (see Figure).

$text{Sol. :}$

$text{Given :}$

$text{Diameter of hemisphere & Cylinder }d = 2.8text{ cm}$

$text{Radius of hemisphere & Cylinder }r = 1.4text{ cm}$

$text{Length of Cylinder }h = 5 - 2(1.4) = 2.2text{ cm}$

$text{To find :}$

$text{Volume of syrup}$ .

$text{Solve :}$

$text{Volume of a gulab jamun = Volume of Cylinder}$

$text{ + 2Volume of hemishere}$

$text{V} = pir^2h + 2times2/3pir^3$

$text{ } = pir^2(h + 2times2/3r)$

$text{ } = 22/7times(1.4)^2(2.2 + 2times2/3times1.4)$

$text{ } = 22/7times1.4times1.4(2.2 + (4times1.4)/3)$

$text{ } = 22times1.4times0.2((6.6 + 5.6)/3)$

$text{ } = 22times0.28times(12.2)/3$

$text{ } = 6.16times12.2/3$

$text{ } = 75.152/3$

$text{Total Volume of 45 gulab jamun} = 45 times 75.152/3$

$text{ } = 15times75.152$

$text{ } = 1127.28text{cm}^3$

$text{Volume of syrup} = 30% times text{Total Volume}$

$text{ } = 30/100 times 1127.28$

$text{ } = 338.184text{cm}^3$

$text{Answer :}$

$text{Volume of syrup 338.184 cm}^3$ .

Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are

$text{15 cm by 10 cm by 3.5 cm.}$ The radius of each of the depressions is

$text{0.5 cm}$ and the depth is

$text{1.4 cm}$ . Find the volume of wood in the entire stand (see figure).

$text{Sol. :}$

$text{Given :}$

$text{Dimensions of the cuboid 15 cm, 10 cm, 3.5 cm}$

$text{Radius of dipressions }r =0.5text{ cm}$

$text{Depth of dipressions }h = 1.4text{ cm}$

$text{To find :}$

$text{The volume of wood in the entire stand.}$

$text{Solve :}$

$text{Volume of wood = Volume of cuboid}$

$text{ - Volume of 4 conical dipression}$

$text{V} = LtimesBtimesH - 4times1/3pir^2h$

$text{} = 15times10times3.5 - 4times1/3times22/7times(0.5)^2times1.4$

$text{} = 525 - 4times1/3times22times0.25times0.2$

$text{} = 525 - 1.47$

$text{} = 523.53text{cm}^3$

$text{Answer :}$

$text{The Volume of wood 523.53 cm}^3.$

Q5. A vessel is in the form of an inverted cone. Its height is

$text{8 cm}$ and the radius of its top, which is open, is

$text{5 cm.}$ It is filled with water up to the brim. When lead shots, each of which is a sphere of radius

$text{0.5 cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

$text{Sol. :}$

$text{Given :}$

$text{Radius of Cone }R = 5text{cm}$

$text{Height of Cone }h = 8text{cm}$

$text{Radius of Sphere }r = 0.5text{cm}$

$text{To find :}$

$text{The number of lead shots dropped in the vessel.}$

$text{Solve :}$

$text{Volume of Conical vessel} = 1/3piR^2h$

$text{V} = 1/3pi(5)^2times8$

$text{ } = 200/3pi$

$text{Volume of water flown out }V_1 = 1/4timesV$

$V_1 = 1/4times200/3pi$

$V_1 = 50/3pi$

$text{The Vol. of n lead shots = Vol. of water displaced}$

$ntimes4/3pi(r)^3 = V_1$

$ntimes4/3pi(0.5)^3 = 50/3pi$

$n = (50times3)/(3times4times0.125)$

$n = (50)/(4times0.125)$

$n = 50/0.5$

$n = 500/5$

$n = 100$

$text{Answer :}$

$text{The number of lead shots in vessel are 100}$ .

Q6. A solid iron poe consists of a cylinder of height

$text{220 cm}$ and base diameter

$text{24 cm},$ which is surmounted by another cylinder of height

$text{60 cm}$ and radius

$text{8 cm}$ . Find mass of the pole, given that

$text{1 cm}^3$ of iron has approximately

$text{8 g}$ mass. (Use

$pi = 3.14$ )

$text{Sol. :}$

$text{Given :}$

$text{Height of Cylindrical pole }h_1 = 220text{ cm}$

$text{Diameter of Cylindrical pole }d= 24text{ cm}$

$text{Radius of Cylindrical pole }r_1 = 12text{ cm}$

$text{Height of another Cylinder }h_2 = 60text{ cm}$

$text{Radius of another Cylinder }r_2 = 8text{ cm}$

$text{To find :}$

$text{The mass of the pole}.$

$text{Solve :}$

$text{Tota Vol. of pole = Vol. of Pole I+Vol. of Pole II}$

$text{V} = pir_1^2h_1 + pir_2^2h_2$

$text{ } = pi(r_1^2h_1 + r_2^2h_2)$

$text{ } = 3.14((12)^2times220 + (8)^2times60)$

$text{ } = 3.14(144times220 + 64times60)$

$text{ } = 3.14(31680 + 3840)$

$text{ } = 3.14(35520)$

$text{ } = 111532.8 text{cm}^3$

$text{mass of iron pole = V}timesRate$

$text{mass of iron pole = }111532.8times 8$

$text{ = 892262.4 g}$

$text{ = 892.26 kg (approx)}$

$text{Answer :}$

$text{The mass of iron pole is 892.26 kg (approx)}.$

Q7. A solid consisting of a right circular cone of height

$text{120 cm}$ and radius

$text{60 cm}$ standing on a hemisphere of radius

$text{60 cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is

$text{60 cm}$ and its height is

$text{180 cm}$ .

$text{Sol. :}$

$text{Given :}$

$text{Height of Cylinder }H = 180text{ cm}$

$text{Radius of Cylinder, Cone & hemi. }r = 60text{ cm}$

$text{Heigth of Cone }h = 180 - 60 = 120text{ cm}$

$text{To find :}$

$text{The Volume of water left in the cylinder}$

$text{Solve :}$

$text{Vol. of water = Vol. of Cylinder}$

$text{ - (Vol. of Cone + Vol. of hemisphere)}$

$text{V} = pir^2H - (1/3pir^2h + 2/3pir^3)$

$text{ } = pir^2[H - 1/3(h + 2r)]$

$text{ } = 22/7(60)^2[180 - 1/3 (120 + 2times60)]$

$text{ } = 22/7times3600[180 - 1/3(120 + 120)]$

$text{ } = 11314.285[180 - 1/3times240]$

$text{ } = 11314.285[180 - 80]$

$text{ } = 11314.285times100$

$text{ } = 1131428.500text{ cm}^3$

$text{ } = 1.131text{ m}^3$

$text{Answer :}$

$text{The Vol. of the water left in Cylinder 1.131 m}^3.$

Q8. A spherical glass vessel has a cylindrical neck

$text{8 cm}$ long,

$text{2 cm}$ in diameter, the diameter of the spherical part is

$text{8.5 cm}$ . By measuring the amount of water it holds, a child finds its volume to be

$text{345 cm}^3$ . Check whether she is correct, taking the above as the inside measurements, and

$pi = 3.14$ .

$text{Sol. :}$

$text{Given :}$

$text{Diameter of spherical part }D = 8.5text{ cm}$

$text{Radius of spherical part }R = 8.5/2text{ cm}$

$text{Diameter of Cylindrical part }d = 2text{ cm}$

$text{Radius of Cylindrical part }r = 1text{ cm}$

text{Height of Cylindrical part }h = 8text{ cm}`

$text{To find :}$

$text{Volume of Vessel}$

$text{Solve :}$

$text{Total Vol. of Vessel = Vol. of Sphere}$

$text{ + Vol. of Cylinder}$

$text{V} = 4/3piR^3 + pir^2h$

$text{ } = pi(4/3R^3 + r^2h)$

$text{ } = 3.14(4/3times(8.5/2)^3 + (1)^2times8)$

$text{ } = 3.14(4/3times76.76 + 8)$

$text{ } = 3.14(102.35 + 8)$

$text{ } = 3.14times110.35$

$text{ } = 346.51text{ cm}^3$

$text{Answer :}$

$text{Volume of vessel is 346.51 cm}^3$ .

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10th Maths 13.2

NCERT Class 10th solution of Exercise 13.1

NCERT Class 10th solution of Exercise 13.3

NCERT Class 10th Maths Projects

Chapter 13

Surface Areas And Volume

Exercise 13.2

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